If \(f(x) = mx^{2} - 6x - 3\) and \(f'(1) = 12\), find the value of the constant m.
\(f(x) = mx^{2} - 6x - 3\)
\(f '(x) = \frac{\mathrm d y}{\mathrm d x} = 2mx - 6\)
\(f'(1) = 2m - 6 = 12\)
\(2m = 18 \implies m = 9\)
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